Input interpretation
![HNO_3 nitric acid + Sn white tin ⟶ H_2O water + NO_2 nitrogen dioxide + H2SnO3](../image_source/8b726986ee333e7401899120c15dad7b.png)
HNO_3 nitric acid + Sn white tin ⟶ H_2O water + NO_2 nitrogen dioxide + H2SnO3
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sn ⟶ c_3 H_2O + c_4 NO_2 + c_5 H2SnO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sn: H: | c_1 = 2 c_3 + 2 c_5 N: | c_1 = c_4 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Sn: | c_2 = c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 1 c_4 = 4 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3](../image_source/85f25ab1eef178af902cd0e40db8d135.png)
Balance the chemical equation algebraically: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sn ⟶ c_3 H_2O + c_4 NO_2 + c_5 H2SnO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sn: H: | c_1 = 2 c_3 + 2 c_5 N: | c_1 = c_4 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Sn: | c_2 = c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 1 c_4 = 4 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3
Structures
![+ ⟶ + + H2SnO3](../image_source/6484fb6d135b1d32ae5407f7d9244690.png)
+ ⟶ + + H2SnO3
Names
![nitric acid + white tin ⟶ water + nitrogen dioxide + H2SnO3](../image_source/d8c363e283e34250ede5f727a4a134dc.png)
nitric acid + white tin ⟶ water + nitrogen dioxide + H2SnO3
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 4 | -4 Sn | 1 | -1 H_2O | 1 | 1 NO_2 | 4 | 4 H2SnO3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 4 | -4 | ([HNO3])^(-4) Sn | 1 | -1 | ([Sn])^(-1) H_2O | 1 | 1 | [H2O] NO_2 | 4 | 4 | ([NO2])^4 H2SnO3 | 1 | 1 | [H2SnO3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HNO3])^(-4) ([Sn])^(-1) [H2O] ([NO2])^4 [H2SnO3] = ([H2O] ([NO2])^4 [H2SnO3])/(([HNO3])^4 [Sn])](../image_source/966af126649141510ea531ac782ff153.png)
Construct the equilibrium constant, K, expression for: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 4 | -4 Sn | 1 | -1 H_2O | 1 | 1 NO_2 | 4 | 4 H2SnO3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 4 | -4 | ([HNO3])^(-4) Sn | 1 | -1 | ([Sn])^(-1) H_2O | 1 | 1 | [H2O] NO_2 | 4 | 4 | ([NO2])^4 H2SnO3 | 1 | 1 | [H2SnO3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HNO3])^(-4) ([Sn])^(-1) [H2O] ([NO2])^4 [H2SnO3] = ([H2O] ([NO2])^4 [H2SnO3])/(([HNO3])^4 [Sn])
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 4 | -4 Sn | 1 | -1 H_2O | 1 | 1 NO_2 | 4 | 4 H2SnO3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Sn | 1 | -1 | -(Δ[Sn])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NO_2 | 4 | 4 | 1/4 (Δ[NO2])/(Δt) H2SnO3 | 1 | 1 | (Δ[H2SnO3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/4 (Δ[HNO3])/(Δt) = -(Δ[Sn])/(Δt) = (Δ[H2O])/(Δt) = 1/4 (Δ[NO2])/(Δt) = (Δ[H2SnO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/6ef4fd924e85e396102c9bfa60e7892b.png)
Construct the rate of reaction expression for: HNO_3 + Sn ⟶ H_2O + NO_2 + H2SnO3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 4 HNO_3 + Sn ⟶ H_2O + 4 NO_2 + H2SnO3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 4 | -4 Sn | 1 | -1 H_2O | 1 | 1 NO_2 | 4 | 4 H2SnO3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Sn | 1 | -1 | -(Δ[Sn])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NO_2 | 4 | 4 | 1/4 (Δ[NO2])/(Δt) H2SnO3 | 1 | 1 | (Δ[H2SnO3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/4 (Δ[HNO3])/(Δt) = -(Δ[Sn])/(Δt) = (Δ[H2O])/(Δt) = 1/4 (Δ[NO2])/(Δt) = (Δ[H2SnO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | white tin | water | nitrogen dioxide | H2SnO3 formula | HNO_3 | Sn | H_2O | NO_2 | H2SnO3 Hill formula | HNO_3 | Sn | H_2O | NO_2 | H2O3Sn name | nitric acid | white tin | water | nitrogen dioxide | IUPAC name | nitric acid | tin | water | Nitrogen dioxide |](../image_source/4c131e4f051cb4796c2cf9edbc540e53.png)
| nitric acid | white tin | water | nitrogen dioxide | H2SnO3 formula | HNO_3 | Sn | H_2O | NO_2 | H2SnO3 Hill formula | HNO_3 | Sn | H_2O | NO_2 | H2O3Sn name | nitric acid | white tin | water | nitrogen dioxide | IUPAC name | nitric acid | tin | water | Nitrogen dioxide |
Substance properties
![| nitric acid | white tin | water | nitrogen dioxide | H2SnO3 molar mass | 63.012 g/mol | 118.71 g/mol | 18.015 g/mol | 46.005 g/mol | 168.72 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 231.9 °C | 0 °C | -11 °C | boiling point | 83 °C | 2602 °C | 99.9839 °C | 21 °C | density | 1.5129 g/cm^3 | 7.31 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | reacts | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 600 °C) | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless | |](../image_source/76160f2773f8bccd48a3277a6a09f4fc.png)
| nitric acid | white tin | water | nitrogen dioxide | H2SnO3 molar mass | 63.012 g/mol | 118.71 g/mol | 18.015 g/mol | 46.005 g/mol | 168.72 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 231.9 °C | 0 °C | -11 °C | boiling point | 83 °C | 2602 °C | 99.9839 °C | 21 °C | density | 1.5129 g/cm^3 | 7.31 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | reacts | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 600 °C) | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless | |
Units